f(x) = (x - 4)² + 2 x(x+5) - 17
1) démontrer que pour tout x, on a
f(x) = 3 x² + 2 x - 1 et f(x) = (3 x - 1)(x+1)
f(x) = (x - 4)² + 2 x(x+5) - 17 il suffit de développer f(x)
f(x) = x² - 8 x + 16 + 2 x² + 10 x - 17
= 3 x² + 2 x - 1
maintenant il faut mettre f(x) sous forme canonique
f(x) = 3 x² + 2 x - 1
forme canonique de f(x) = a(x - α)²+ β
α = - b/2a = - 2/6 = - 1/3
β = f(α) = f(- 1/3) = 3 (-1/3)² + 2(-1/3) - 1 = 1/3 - 2/3 - 1 = - 1/3 - 1 = - 4/3
f(x) = 3(x + 1/3)² - 4/3 = 1/3[(3 x + 1)² - 4]
f(x) = 1/3(3 x + 1 + 2)(3 x + 1 - 2) = 1/3(3 x + 3)(3 x - 1) = (x + 1)(3 x - 1)
